Question

A random sample of 30 receipts for individuals shopping at the Community Drug Store showed the sample mean to be`x = $28. 19 with sample standard deviation s = $4.06 a. Use Chevyshev’s theorem to find the smallest interval centered on the mean in which we can expect at least 75% of the data to fall.

Answer 1

Chebyshev's inequality states that

$\mathbb P(|X-\mu|\ge k\sigma)\le\dfrac1{k^2}\implies\mathbb P(|X-\mu|\le k\sigma)\ge1-\dfrac1{k^2}$

We assume $\mu=\bar x=28.19$ and $\sigma=s=4.06$, so that the inequality is

$\mathbb P(|X-28.19|\le4.06k)\ge1-\dfrac1{k^2}$

and since we want the probability to be at least 0.75, we have

$1-\dfrac1{k^2}=\dfrac34\implies k=2$

which means we get from Chebyshev's inequality that

$\mathbb P(|X-28.19|\le8.12)=\mathbb P(20.07\le X\le36.31)\ge0.75$